Problem: Evaluate $\int\dfrac{2x^3-3x^2-3x+2}{x-2}\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{2x^3}3+\dfrac{x^2}2-x+C$ (Choice B) B $\dfrac{2x^3}3-\dfrac{x^2}2+x+\ln|x-2|+C$ (Choice C) C $\dfrac{2x^3}3+\dfrac{5x^2}2-x+C$ (Choice D) D $\dfrac{2x^3}3-\dfrac{x^2}2-x+\ln|x-2|+C$
Explanation: We can rewrite the integrand using polynomial long division, because the degree of the numerator is greater than or equal to the degree of the denominator. After performing the division, we obtain the following result: $\dfrac{2x^3-3x^2-3x+2}{x-2}=2x^2+x-1$ Now we can perform the integration: $\begin{aligned} &\phantom{=}\int\dfrac{2x^3-3x^2-3x+2}{x-2}\,dx \\\\ &=\int\left(2x^2+x-1\right)\,dx \\\\ &=2\int x^2\,dx+\int x\,dx-\int 1\,dx \\\\ &=2\dfrac{x^3}{3}+\dfrac{x^2}{2}-x+C \\\\ &=\dfrac{2x^3}3+\dfrac{x^2}2-x+C \end{aligned}$ In conclusion, $\int\dfrac{2x^3-3x^2-3x+2}{x-2}\,dx=\dfrac{2x^3}3+\dfrac{x^2}2-x+C$